Question: Find $\lim_{x\to 3}\dfrac{x-3}{\sqrt{4x+4}-4}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-4$ (Choice B) B $1$ (Choice C) C $2$ (Choice D) D The limit doesn't exist
Solution: Substituting $x=3$ into $\dfrac{x-3}{\sqrt{4x+4}-4}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression with a square root on our hands, let's try to re-write it using the method of rationalization. $\begin{aligned} &\phantom{=}\dfrac{x-3}{\sqrt{4x+4}-4} \\\\ &=\dfrac{x-3}{\sqrt{4x+4}-4}\cdot\dfrac{\sqrt{4x+4}+4}{\sqrt{4x+4}+4} \gray{\text{Rationalize the denominator}} \\\\ &=\dfrac{(x-3)(\sqrt{4x+4}+4)}{(4x+4)-4^2} \\\\ &=\dfrac{\cancel{(x-3)}(\sqrt{4x+4}+4)}{4\cancel{(x-3)}} \gray{\text{Cancel out common factors}} \\\\ &=\dfrac{\sqrt{4x+4}+4}{4} \text{, for }x\neq 3 \end{aligned}$ This means that the two expressions have the same value for all $x$ -values (in their domains) except for $3$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{x-3}{\sqrt{4x+4}-4}=\dfrac{\sqrt{4x+4}+4}{4}$ for all $x$ -values in the interval $(2.5,3.5)$ except for $x=3$. Therefore, $\lim_{x\to 3}\dfrac{x-3}{\sqrt{4x+4}-4}=\lim_{x\to 3}\dfrac{\sqrt{4x+4}+4}{4}=2$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to 3}\dfrac{x-3}{\sqrt{4x+4}-4}=2$.